[Math] If $n \times n$ matrix $M$ is skew-symmetric and invertible, prove that $n$ is even

Question

I am asked the following:

Assume that $M$ is skew-symmetric and invertible. If $M$ is an $n \times n$ matrix, prove that $n$ is even.

The following is my attempt at the solution. I'm not too sure if it is right, or if I'm allowed to do this.

So, if a matrix is skew-symmetric, then,

$$A^T=-A$$

and by taking the determinant of both sides,
$$\det(A^T)=\det(-A)$$

Using the property of transpose and the multi-linear property,

$$\det(A)=(-1)^n\det(A)$$

Here is where my problem arises. Are we allowed to assume that $n$ is even and then show that it is invertible? However, if we do this we get

$$\det(A) = \det(A)$$

and I'm not too sure what this equality means. Or where to go from here. Any help would be greatly appreciated.

Answer 1

Note that if $n$ is odd, you would get $\det (A) = - \det (A)$, so A has determinant $0$ and is not invertible. Some invertible matrices that are even by even are skew-symmetric; try constructing them yourself!